• # question_answer An emf $E=100\,sin\,314\,t$ is applied across a pure capacitor of $637\mu F.$ Find (i) the instantaneous current I (ii) instantaneous power P (iii) the frequency of power and (iv) the maximum energy stored in the capacitor.

(i) Given, $E=100\,sin\,314\,t\,volt$ As the current in a capacitor leads the voltage by $90{}^\circ ,$ so the instantaneous current is given by             $I={{I}_{0}}\sin (314\,t+90{}^\circ )={{I}_{0}}\cos 31\,4t$ where,   ${{I}_{0}}=\frac{{{E}_{0}}}{{{X}_{C}}}=\frac{{{E}_{0}}}{1/\varepsilon C}={{E}_{0}}\omega C$ But, ${{E}_{0}}=100V,$ $\omega =314\,rad\,{{s}^{-1}},$ $C=637\times {{10}^{-6}}F$ $\therefore$      ${{I}_{0}}=100\times 314\times 637\times {{10}^{-6}}=20A$ Hence,  $I=20cos\text{ }314\,t$ ampere. (ii) Instantaneous power,             $P=EI=100\sin 314\,t\times 20\cos 314\,t$             $=1000\sin 628\,t\,watt$ (iii) Angular frequency of power, ${{\omega }_{p}}=628\,rad\,{{s}^{-1}}$ $\therefore$ Frequency of power, ${{f}_{H}}=\frac{{{\omega }_{H}}}{2\pi }=\frac{628}{2\pi }=100Hz$ (iv) The maximum energy stored in the capacitor is ${{U}_{0}}=\frac{1}{2}CE_{0}^{2}=\frac{1}{2}\times 637\times {{10}^{-6}}\times {{(100)}^{2}}=3.185\,J$